Physics midterms
A 5.3-kg steel ball is dropped from a height of 15.0m into a box of sand and sinks 0.70 m into the sand before stopping. How much energy is dissipated through the interaction with the sand?
energy dissipated= ∆U= mgh
In terms of variables given and fundamental constants, how much does the internal energy of the system of cart and bullet change?
∆Eint= -∆K ---> ∆Eint= Ki -Kf ----> add in what we have
Draw a vector indicating the direction of Δv the change in velocity of the block between times t4 and t10. If Δv is zero, state so explicitly. Explain how you used the strobe diagram to determine your answer.
-v4 is pointing left, v10 is pointing right ∆v= v10-v4 --> v10= v4 + ∆v therefore ∆v must point right
Lab: Given a graph for a rotating wheel at a constant speed: 1) What is the speed in the x-direction after 1 rotation? 2) What is the speed in the y-direction? 3) What is the angular acceleration? 4) what is the centripital acceleration
1) 0 m/s, because this is the turn around point for x 2) v= wr and w= 2pi/t 3) a= ∆w/∆t. since it is moving at a constant speed, this is 0 4) Ac= v^2/r
Which choice best represents the energy dissipated when: 1) a car traveling with kinetic energy K has a totally inelastic collision with an identical, stationary car? 2) a car traveling with kinetic energy K has a totally inelastic collision with a fixed concrete wall?
1) Ki= 1/2mv^2. When the collision occurs, the speed is halved, and the inertia is doubled, therefore Kf= 1/2(2m)(1/2v)^2 or 1/4mv^2. therefore K/2 is converted 2) since Kf=0, all of K is converted here
If there were a small amount of friction between the cart and track, what is true?
1) the time rate of change momentum is less than w/ no friction because p=mv and v decreases with friction, therefore p does as well 2) Tension goes up. T= mg-ma and a decreases
Relative to earth a person is moving to the right at 𝑣𝑝. In terms of variables given and fundamental constants, what does she observe as the velocity of the cart-bullet system after the collision?
Vobserved = Vf -Vp
A ball of mass 2kg moving horizontally with speed 67m/s hits a wall and rebounds with a speed of 52m/s. If the ball is in contact with the wall for 2.5 milliseconds, what is the average force exerted by the ball on the wall?
F= ∆p/∆t make sure to make one speed negative because of the direction change
As shown, block 1 is traveling to the right at 1.8 m/s. It collides with a block 2 traveling to the left at 0.2 m/s. After the collision block 1 travels to the left at 0.6 m/s and block 2 travels to the right at 1.4 m/s. The magnitude of the change in momentum of block 1 is 9.6 kg.m/s, and the collision occurs over 8.0×10−3s. During the collision what is the size of the average force of block 2 on block 1?
Favg= ∆p / ∆t. both values are given
What is the center of mass acceleration of the system consisting of both blocks and the spring?
Fext= Msystem x Acm
A 4.00-kg block rests between the floor and a 3.00-kg block as shown in the figure. The 3.00-kg block is tied to a wall by a horizontal rope. If the coefficient of static friction is 0.800 between each pair of surfaces in contact, what horizontal force F must be applied to the 4.00-kg block to make it move?
First there is the gravitational force of both blocks, (3+4)g so multiply this by the coefficient of static friction, then there's the static friction of the 3kg block so 3gµ then add them together
A person in an apartment below notices that it takes 0.17 s for the flower pot to fall past his 1.9 m tall window. The roof of the apartment building is 4.95 m above the top of the window. What does the person determine the vertical velocity of the flowerpot was as it left the top of the roof?
First use Xf=Xi + Vo∆t +1/2a∆t^2 with the info from the person to solve for Vo, which can be used as the Vf in the equation Vf^2= Vi^2 + 2a∆x and solve for Vo
What is the normal force at the top of a loop the loop?
Fnet=mg+Fn Fnet=ma=m(v^2/r) = mg +Fn so Fn=m(v^2/r)-mg
For the collision, is the magnitude of the acceleration of the bullet, greater than, less than, or the same as the magnitude of the acceleration of the cart? Explain.
Interaction pair so forces are equal. This means ma=ma since the cart has a higher mass, the bullet must have a higher acceleration
Draw an energy diagram of a system of a person, both block, and the spring vs. just the blocks and spring
Just blocks and spring: Some ∆K, a little ∆U, No ∆Esource or ∆Ethermal. W=∆K+∆U Person,blocks,spring: ∆K and ∆U same as above, but also a little amount of ∆Ethermal, and a negative ∆Esource equal to ∆K+∆U+∆Ethermal because in a closed system ∆E=0, and W=∆E so W must =0
Is the magnitude of the force by block C on block D in case 2 greater than, less than, or equal to the magnitude of the force by block C on block B in case 1? Explain.
Less than because the acceleration of block C decreases and the inertia of block C is the same.
A 100-kg person runs off the end of a horizontal pier and lands on a free floating 200-kg raft, which was initially at rest. After the person lands on the raft, the raft with person on it moves away from the pier at 1 m/s. What was this person's speed as they jumped off the pier?
MpVi= (Mp+Mr)Vf
For the collision, is the magnitude of the change in momentum of the bullet, greater than, less than, or the same as the magnitude of the change in momentum of the cart? Explain
No external forces mean that the system is isolated, and therefore momentum must be conserved. This means the change in momentum is equal.
Block B is now replaced by block D, which has an inertia much greater than the inertia of block B (mD > mB). The hand is still pushing with the same force. Call this situation case 2. Is the net force on block A the same, more, or less?
Since the system inertia increases, the system acceleration must decrease because the force is equal. This means the net force will be less as the acceleration decreases
What is the inertia of the cart? (m1)
T= m1a ---> m1=T/a
If the table now had friction what would happen to the tension?
Tension would increase
A block with an initial speed of 4.5m/s goes up incline of 30º. How far up the incline does it go?
Vf= Vi + a∆t. a= gsinø Vf=0, solve for ∆t then Xf=Xi + Vo∆t + 1/2a∆t
A 5.00-kg object slides down a frictionless surface inclined at an angle of 30.0° from the horizontal. The total distance moved by the object along the inclined surface is 10.0 meters. The work done on the object by the force of gravity is
W= F∆x since we are looking for work done by gravity, we need the ∆y not ∆x so we take sin(30)x10m =5m F=mg so W=mg∆y
Two identical blocks, each with inertia 75.0 kg, are connected by a massless spring. The blocks are at rest and there is no friction between the blocks and the surface. As shown, you then push the left block with a constant force of 8.00 N to the right. After you have pushed the left block by 0.500 m the kinetic energy of the system consisting of both blocks and the spring is 3 J, and the spring has been compressed by 0.050 m What is the spring constant?
W=∆E=∆K+∆U ∆U= 1/2k(X-Xo)^2 solve
How far off the ground is the hanging object at the start of the experiment?
Xf=Xi + Vo∆t + 1/2a∆t^2 (Vo=0)
A object with inertia of 2.0 kg is moving in 1- dimension with positive 𝑥 velocity. The potential energy curve as a function of position is shown in the figure. The kinetic energy of the object at the origin is 12 J. The system is conservative, and there is no friction. What will be the kinetic energy when the inertia is at 𝑥 = 2.0 m?
add 12J to the initial energy, then go to x=2m and Etotal must be the same
Arthur and betty start walking toward each other from 100m apart, with Arthur walking at 3m/s and betty at 2m/s. How far will Betty have walked when they meet and how much time will have passed?
arthur walks farther than betty by a ratio of 3/2. so Betty walked 40m. at 2m/s this will take her 20s
Rank the inertia of the carts A, B, and S based on the following collision diagrams.
assess the ∆v for each cart, and the lower ∆v means a greater inertia
Is the magnitude of the average acceleration of the block between times t4 and t7 greater than, less than, or equal to the magnitude of the average acceleration of the block between times t4 and t10? Explain
average acceleration= ∆v / ∆t since the speeds are equal, ∆v4-7 = 2(∆v4-10) ∆t4-7= 2(∆v4-10) average accelerations are equal
Is the magnitude of the average velocity of the block between times t2 and t4 greater than, less than, or equal to the magnitude of the average velocity of the block between times t2 and t10? Explain how you used the strobe diagram to determine your answer.
because the displacement is the same from t2-4 and t2-10, we can focus on the ∆t since average velocity= ∆x/∆t. t2-10 obviously has a higher ∆t therefore a smaller average velocity.
When does the particle change direction and which way does it start going? (equation given)
derive to v(x), then set equal to 0 and solve for t (because at the turn around point v=0. then draw the graph and look at it
average speed?
distance/∆t
Given that there are samples weighing 0.021kg on each side of the 0.2m rod and the total rotational inertia is 1.1x10^-3, what is the rotational inertia of the rod in the centrifuge?
find the I of one sample using the radius (0.2m/2) and the mass, then multiply this number by 2 since there are 2 samples, then subtract from the overall I value and solve for mass using 0.2m as the rods radius since the entire thing is rotating.
An 80kg man sits on a calm lake in a 50kg boat. the boat's seats are 4m apart and the closest seat is 2m from the dock. The man walks from the furthest end (6m) to the closest end (2m). how far is the man now from the dock?
he walks 4m, and the ratio of weight is 50/80 so we multiply this by 4m to see it moves 2.5m. Since it was originally 2m away the man is now 4.5m away from the boat
A ball with inertia 0.5kg is moving with v1= 4.5i+(-6.5)j before it collides with the ground and v2= 4.5i+5.5j after. What is the magnitude of impulse from the ground on the ball?
impulse=J=∆p= m∆v v2-v1= 12m/s x 0.5kg= 6N/s
A box of mass 𝑚2 slides on a frictionless horizontal table pulled by a string attached to a hanging weight of mass 𝑚1 as shown. The pulley over which the string is strung is massless and frictionless. What is the tension in the string?
m1a = m1g - T T=m2a solve both for a and set equal to each other, then solve for T
A cart of inertia 𝑚1 is attached to a string hung over a pulley. A hanging object with inertia, 𝑚2, is attached at the other end of the string. A force sensor on the cart records the force applied to the cart by the string. Data taking is started at 𝑡 = 0 s and the cart is held at rest and is then released at 𝑡 =0.4 s. The average acceleration measured between 0.4 s and 1.4 s is 2.0 m/s2. What is the inertia of the hanger? (m2)
m2g - T= m2a ----> m2g - m2a = T ---> m2(g-a)=T m2= T/(g-a)
Which choice best represents the position of the center of mass of the 8 remaining blocks?
measure the mass as the number of blocks in each chunk and the position as the number of blocks over from the left. The plug into Rcm equation
As shown a bullet with inertia 𝑚𝑏 is fired at a cart with inertia 𝑚𝑐. Note that 𝑚𝑏 < 𝑚𝑐. After the collision, the cart rolls at 𝑣𝑓 relative to earth. In terms of variables given and fundamental constants, what is the bullet's speed relative to earth before it strikes the cart?
momentum is conserved as it's a completely inelastic collision. MbVi= (Mb+Mc)Vf ---> Vi= ((Mb+Mc)Vf) /Mb
Which expression predicts the time rate of change momentum for the system of the cart?
remember that ∆p/∆t=F and in this case F=Tension. So we can solve for tension as we did before, writing an equation for each cart, solve for a, set equal, and solve for T
Blocks A, B, and C are being pushed to the right across a frictionless table by a hand applying a constant horizontal force as shown. The inertias of blocks A, B, and C are 2M, 3M, and M, respectively (i.e., mC < mA < mB). Draw net force vectors for each block
since F=ma and all blocks are accelerating together, the forces are equal to the magnitude of the masses
A car goes into a skid and gradually comes to a stop, accelerating at a constant rate. At the midpoint of the skid (in terms of position), how much of its kinetic energy has it lost?
solve Vf^2= Vi^2 + 2a∆x first for ∆x and then plug this value in for Vf^2= Vi^2 + 2a(∆x/2) and eliminate a, then solve for Vf. plug this into K=1/2mv^2 and compare to a normal 1/2mv^2
A centrifuge rotates at 150 revs/s, and at t=0 it slows down with an angular acceleration of 9.4s^-2. How many revolutions does the centrifuge make before it stops?
solve for ∆ø, remembering to multiply 150revs/s x2pi. then you have the distance it went, so divide by 2pi to get the revolutions
Which swimmer gets across first?
swimmer that swims straight across
Lab: What distance has the cart traveled from the beginning of the recording to when it comes in contact with the stop?
take the average velocity for the time the cart was moving (Vf/2) then multiply by the amount of time the cart was moving
A book is at rest on a surface with an incline of 15º. What is the interaction pair of the gravitational force exerted on the book by earth?
the gravitational force exerted by the book on earth
You take a uniform circle with radius 2𝑟 and cut out a concentric circle of radius 𝑟. As shown you place the cut out circle to the left of the bigger circle that now has a hole in it. Where is the center of mass of this system of objects?
use (pi)r^2 for mass, and dont forget to subract the cut out circle
A block has an initial speed of 7.0 m/s up an inclined plane that makes an angle of 32° with the horizontal. Ignoring friction, what is the block's speed after it has traveled 2.0 m?
use equation a= gsinø to find acceleration, then we have a, Vo, and ∆x so plug into Vf^2= Vi^2 + 2a∆x and solve for Vf
A bullet moving at Vi strikes a block of length L. Which expression is a good estimate of the time interval from when the bullet first hits the block until it arrives at the center of the block?
v=x/t ---> t=x/v x= L/2 Vavg= Vi/2 ∆t=L/Vi
What is spring constant k given Vi, mass, Vf, and the distance the spring was compressed?
we know an equation for ∆Uspring, and in this case ∆U=∆K, which starts with the Vf have(now Vi). We're interested in the part where Vf=0 so ∆Uspring=1/2mVi^2 then plug that value into the equation
a projectile is launched at a speed of 39.1 m/s at an angle of 25º and it lands on a target 120m away. How long was the projectile in the air?
we need the x component of velocity so 39.1(cos(25))=v and then x/v=t
Runners A and B run a 100-m race, each at a constant speed. Runner A takes first place, beating runner B by 10 m. By what time interval does runner A beat runner B?
write equations for Va/Vb= in terms of distance travelled for a (should get a fraction) and in terms of the extra time needed by b to get the full 100m (∆ta+∆t). then set those 2 equal to each other and solve.
Now consider the same situation but there is a friction force of 2.00 N between the left block and the surface. We still push with a 8.00 N force to the right and the left block moves 0.500 m. Would the thermal energy of the system consisting of both blocks and the spring increase by more than 1.0 J, exactly 1.0 J, less than 1.0 J, or the thermal energy would not change? Explain.
∆Ethermal= Work done by Friction= Ffrict∆x 2x.5= 1J, but not all of this goes into the system, some goes into surface. therefore the answer is less than 1J
